In the previous section we saw two solved examples on grouped frequency distribution tables. In this section, we will see more solved examples.
Solved example 25.3
The relative humidity (in %) of a certain city for a month of 30 days was as follows:
(i) Construct a grouped frequency distribution table with classes 84 - 86, 86 - 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
Part (i)
1. The smallest value is 84.9. So the first interval can be 84-86.
2. It is given that the classes can be 84-86, 86-88 etc., So the class width is the same, which is equal to 2. Thus the classes that we use in this problem are: 84-86, 86-88, 88-90, 90-92, ... , 98-100
3. The largest value is 99.2. So 98-100 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
Sample calculation:
• Consider the class-interval 94-96. The values from 94 to 96 in the raw data are:
95.3, 94.2, 95.1, 95.1, 95.2, 95.7
• All together, there are 6 values
• Thus, the frequency of the class-interval 94-96 is 6
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
The relative humidity is high on all days of the given month. So the recordings are made in a rainy season
Part (iii)
The smallest value is 84.9. The largest value is 99.2. So the range is 99.2 -84.9 = 14.3
Solved example 25.4
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
(i) Represent the data given above by a grouped frequency distribution table, taking
the class intervals as 160 - 165, 165 - 170, etc.
(ii) What can you conclude about their heights from the table?
Solution:
Part (i)
1. It is given that the classes can be 160 - 165, 165 - 170 etc., So the class width is the same, which is equal to 5.
2. The smallest value is 150. So the first interval will be 150-155.
2. Thus the classes that we use in this problem are: 150-155, 155-160, ...
3. The largest value is 173. So 170-175 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
Sample calculation:
• Consider the class-interval 155-160. The heights from 155 to 160 in the raw data are:
158, 156, 160, 159, 156, 158, 160, 159, 159, 158, 159
• All together, there are 11 heights. But 160 cm should be included in the next upper class 160-165. There are two 160 values.
• So only 9 distances are eligible to be included in the class-interval 155-160
• Thus, the frequency of the class-interval 155-160 is 9
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
• The heights range from 150 to 172. So an approximate average will be (150+172)/2 = 322/2 = 161
• Let this be rounded to 165 cm
• Then the number of students whose height is equal to or greater than the average 165 cm = 10 +5 = 15
• 15 is less than half of the total 50. So we can say this:
Less than 50% of the students have a height equal to or greater than 165 cm
Solved example 25.5
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
Solution:
Part (i)
1. It is given that the classes can be 0.00 - 0.04, 0.04 - 0.08 etc., So the class width is the same, which is equal to 0.04
2. The smallest value is 0.01. So the first interval will be 0.00 - 0.04.
2. Thus the classes that we use in this problem are: 0.00 - 0.04, 0.04 - 0.08, ...
3. The largest value is 0.22. So 0.20-0.24 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
Sample calculation:
• Consider the class-interval 0.12-0.16. The concentrations from 0.12 to 0.16 in the raw data are:
0.16, 0.12 and 0.13
• All together, there are 3 values. But 0.16 should be included in the next upper class 0.16-0.20
• So only 2 values are eligible to be included in the class-interval 0.12-0.16
• Thus, the frequency of the class-interval 0.12-0.16 is 2
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
1. The concentration greater than 0.11 is 0.12
2. A value of 0.12 will be included in the class 0.12-0.16
3. It will not be included in the class 0.08-0.12
4. So we need to consider only the last three classes in the table
5. The answer is 2 +4 +2 = 8 days
Solved example 25.6
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
Prepare a frequency distribution table for the data given above.
Solution:
In this problem, we do not have to fix up the classes. Because we are not asked to make a 'grouped frequency distribution table'. What we are asked is a 'frequency distribution table'. It is shown below:
Sample calculation:
The number '2' occurs 9 times in the raw data. So the frequency corresponding to '2' is 9
Solved example 25.7
The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digit
Solution:
Let us first separate the 50 digits that occur after the decimal place. The separated digits are written in a tabular form below:
In this problem, we do not have to fix up the classes. Because we are not asked to make a 'grouped frequency distribution table'. What we are asked is a 'frequency distribution table'. It is shown below:
• From the table we can see that 3 and 9 are the most frequently occurring digits
• Also 0 is the least frequently occurring digit.
Solved example 25.8
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 - 10.
(ii) How many children watched television for 15 or more hours a week?
Solution:
Part (i)
1. It is given that one of the classes can be 5-10, and the class width is to be 5
2. The smallest value is 1 hour. So the first interval will be 0-5
2. Thus the classes that we use in this problem are: 0-5, 5-10, ...
3. The largest value is 17. So 15-20 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
Sample calculation:
• Consider the class-interval 10-15. The hours from 10 to 15 in the raw data are:
12, 10, 12, 15, 14 and 12
• All together, there are 6 values. But 15 should be included in the next upper class 15-20
• So only 5 values are eligible to be included in the class-interval 10-15
• Thus, the frequency of the class-interval 10-15 is 5
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
All the values equal to or greater than 15 will be in the last class 15-20. So the required answer is 2
Solved example 25.9
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5
Solution:
Part (i)
1. It is given that the first class is to be 2-2.5, and the class width is to be 0.5
2. The smallest value is 2.2. So the first interval of 2-2.5 is appropriate
2. Thus the classes that we use in this problem are: 2-2.5, 2.5-3.0, ...
3. The largest value is 4.6. So 4.5-5.0 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
Sample calculation:
• Consider the class-interval 4-4.5. The values from 4 to 4.5 in the raw data are:
4.1, 4.5, 4.4, 4.3 and 4.2
• All together, there are 5 values. But 4.5 should be included in the next upper class 4.5-5.0
• So only 4 values are eligible to be included in the class-interval 4.5-5.0
• Thus, the frequency of the class-interval 4.5-5.0 is 4
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
In the next section we will see graphs.
Solved example 25.3
The relative humidity (in %) of a certain city for a month of 30 days was as follows:
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
Part (i)
1. The smallest value is 84.9. So the first interval can be 84-86.
2. It is given that the classes can be 84-86, 86-88 etc., So the class width is the same, which is equal to 2. Thus the classes that we use in this problem are: 84-86, 86-88, 88-90, 90-92, ... , 98-100
3. The largest value is 99.2. So 98-100 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
Sample calculation:
• Consider the class-interval 94-96. The values from 94 to 96 in the raw data are:
95.3, 94.2, 95.1, 95.1, 95.2, 95.7
• All together, there are 6 values
• Thus, the frequency of the class-interval 94-96 is 6
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
The relative humidity is high on all days of the given month. So the recordings are made in a rainy season
Part (iii)
The smallest value is 84.9. The largest value is 99.2. So the range is 99.2 -84.9 = 14.3
Solved example 25.4
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
the class intervals as 160 - 165, 165 - 170, etc.
(ii) What can you conclude about their heights from the table?
Solution:
Part (i)
1. It is given that the classes can be 160 - 165, 165 - 170 etc., So the class width is the same, which is equal to 5.
2. The smallest value is 150. So the first interval will be 150-155.
2. Thus the classes that we use in this problem are: 150-155, 155-160, ...
3. The largest value is 173. So 170-175 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
• Consider the class-interval 155-160. The heights from 155 to 160 in the raw data are:
158, 156, 160, 159, 156, 158, 160, 159, 159, 158, 159
• All together, there are 11 heights. But 160 cm should be included in the next upper class 160-165. There are two 160 values.
• So only 9 distances are eligible to be included in the class-interval 155-160
• Thus, the frequency of the class-interval 155-160 is 9
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
• The heights range from 150 to 172. So an approximate average will be (150+172)/2 = 322/2 = 161
• Let this be rounded to 165 cm
• Then the number of students whose height is equal to or greater than the average 165 cm = 10 +5 = 15
• 15 is less than half of the total 50. So we can say this:
Less than 50% of the students have a height equal to or greater than 165 cm
Solved example 25.5
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
Solution:
Part (i)
1. It is given that the classes can be 0.00 - 0.04, 0.04 - 0.08 etc., So the class width is the same, which is equal to 0.04
2. The smallest value is 0.01. So the first interval will be 0.00 - 0.04.
2. Thus the classes that we use in this problem are: 0.00 - 0.04, 0.04 - 0.08, ...
3. The largest value is 0.22. So 0.20-0.24 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
• Consider the class-interval 0.12-0.16. The concentrations from 0.12 to 0.16 in the raw data are:
0.16, 0.12 and 0.13
• All together, there are 3 values. But 0.16 should be included in the next upper class 0.16-0.20
• So only 2 values are eligible to be included in the class-interval 0.12-0.16
• Thus, the frequency of the class-interval 0.12-0.16 is 2
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
1. The concentration greater than 0.11 is 0.12
2. A value of 0.12 will be included in the class 0.12-0.16
3. It will not be included in the class 0.08-0.12
4. So we need to consider only the last three classes in the table
5. The answer is 2 +4 +2 = 8 days
Solved example 25.6
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
Prepare a frequency distribution table for the data given above.
Solution:
In this problem, we do not have to fix up the classes. Because we are not asked to make a 'grouped frequency distribution table'. What we are asked is a 'frequency distribution table'. It is shown below:
The number '2' occurs 9 times in the raw data. So the frequency corresponding to '2' is 9
Solved example 25.7
The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digit
Solution:
Let us first separate the 50 digits that occur after the decimal place. The separated digits are written in a tabular form below:
• From the table we can see that 3 and 9 are the most frequently occurring digits
• Also 0 is the least frequently occurring digit.
Solved example 25.8
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
(ii) How many children watched television for 15 or more hours a week?
Solution:
Part (i)
1. It is given that one of the classes can be 5-10, and the class width is to be 5
2. The smallest value is 1 hour. So the first interval will be 0-5
2. Thus the classes that we use in this problem are: 0-5, 5-10, ...
3. The largest value is 17. So 15-20 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
• Consider the class-interval 10-15. The hours from 10 to 15 in the raw data are:
12, 10, 12, 15, 14 and 12
• All together, there are 6 values. But 15 should be included in the next upper class 15-20
• So only 5 values are eligible to be included in the class-interval 10-15
• Thus, the frequency of the class-interval 10-15 is 5
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
All the values equal to or greater than 15 will be in the last class 15-20. So the required answer is 2
Solved example 25.9
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5
Solution:
Part (i)
1. It is given that the first class is to be 2-2.5, and the class width is to be 0.5
2. The smallest value is 2.2. So the first interval of 2-2.5 is appropriate
2. Thus the classes that we use in this problem are: 2-2.5, 2.5-3.0, ...
3. The largest value is 4.6. So 4.5-5.0 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:
• Consider the class-interval 4-4.5. The values from 4 to 4.5 in the raw data are:
4.1, 4.5, 4.4, 4.3 and 4.2
• All together, there are 5 values. But 4.5 should be included in the next upper class 4.5-5.0
• So only 4 values are eligible to be included in the class-interval 4.5-5.0
• Thus, the frequency of the class-interval 4.5-5.0 is 4
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
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